= Solution
Use $dz=d^3x\,d^3v$ and set
$$
\rho(\mathbf x,t)=\boxed{\int f(\mathbf x,\mathbf v,t)\,d^3v},\quad
\mathcal E=\frac12\int|\mathbf v|^2f\,dz,\quad
\mathcal H=\int f\log f\,dz.
$$
For unit particle <mass>, $\rho$ is both the <number density> and the <mass density> in the number convention of <1/c>. A fixed reference scale inside the <logarithm> changes $\mathcal H$ by a constant multiple of the conserved total <mass>, leaving its derivative unchanged.
\b[Both prefactors in the source's part (d) need correction.] Multiplication of the <mean-field alignment equation> by $|\mathbf v|^2/2$ and <integration by parts> gives
$$
\dot{\mathcal E}=\int f\mathbf v\cdot\mathbf L[f]\,dz
=\iint K(\mathbf x,\mathbf x_\star)\mathbf v\cdot(\mathbf v_\star-\mathbf v)ff_\star\,dz\,dz_\star.
$$
Exchange $z$ and $z_\star$, use the <symmetry> of $K$, and average the two expressions:
$$
\boxed{\dot{\mathcal E}=-\frac12\iint K(\mathbf x,\mathbf x_\star)|\mathbf v_\star-\mathbf v|^2ff_\star\,dz\,dz_\star\leq0.}
$$
The printed right-hand side omits $1/2$. This corrected <velocity-alignment energy dissipation> agrees with the pair-counting calculation in <1/a>.
For <velocity-alignment entropy production>, let $A(\mathbf x,t)=\int K(\mathbf x,\mathbf x_\star)\rho(\mathbf x_\star,t)\,d^3x_\star$. The <velocity-space divergence> is $\nabla_{\mathbf v}\cdot\mathbf L[f]=-3A$. Assuming $f>0$ and sufficient decay, <integration by parts> yields
$$
\begin{aligned}
\dot{\mathcal H}
&=-\int(1+\log f)\left[\mathbf v\cdot\nabla_{\mathbf x}f+\nabla_{\mathbf v}\cdot(f\mathbf L[f])\right]dz\\
&=\int\mathbf L[f]\cdot\nabla_{\mathbf v}f\,dz
=-\int f\nabla_{\mathbf v}\cdot\mathbf L[f]\,dz\\
&=\boxed{3\iint K(\mathbf x,\mathbf x_\star)\rho(\mathbf x,t)\rho(\mathbf x_\star,t)\,d^3x\,d^3x_\star\geq0.}
\end{aligned}
$$
The printed formula omits the factor $3$ from the three <velocity> components. In $d$ spatial <dimensions>, that factor would be $d$. Zeros of $f$ can be treated by a regularized <entropy> calculation when the required integrability holds.
A concrete smooth counterexample to both printed coefficients is $K\equiv1$ and $f(z,0)=g(\mathbf x)(2\pi)^{-3/2}e^{-|\mathbf v|^2/2}$, where $g$ is a positive normalized <Gaussian distribution>. The total <mass> is one, the total <momentum> is zero, and $\mathbf L[f]=-\mathbf v$ initially. The <kinetic energy> derivative is $-\int |\mathbf v|^2f\,dz=-3$, whereas the printed formula gives $-6$. The <entropy functional> derivative is $3$, whereas the printed formula gives $1$.
The signs describe cooling of relative <velocities> and contraction in <phase space>: $\mathcal H$ increases, so the fine-grained <Gibbs entropy> $-\mathcal H$ decreases. These dissipative particle dynamics are not an isolated <Hamiltonian system>; there is no contradiction with <Liouville's theorem> or the <Boltzmann equation>'s usual <entropy> increase. The microscopic mechanism removing <kinetic energy> is outside this alignment model.
Back to article page
